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Multiple Positive Solutions Of A Singular Fractional Boundary Value Problem

Jiafa Xu and Zhilin Yang

Received 24 November 2009

Abstract

In this paper, we study the existence and multiplicity of positive solutions for the singular fractional boundary value problem

(Dα0+u(t) =h(t)f(t, u(t)),0< t <1, u(0) =u(1) =u0(0) =u0(1) = 0,

where 3 < α ≤ 4 is a real number, Dα0+ is the standard Riemann-Liouville derivative, andh∈C(0,1)∩L(0,1) is nonnegative and may be singular att= 0 and/or t = 1. We use fixed point index theory to establish our main results based on a priori estimates achieved by developing some spectral properties of associated linear integral operators. Our main results essentially extend and improve the corresponding ones in the literature.

1 Introduction

In this paper, we study the existence and multiplicity of positive solutions for the singular fractional boundary value problem

(Dα0+u(t) =h(t)f(t, u(t)), t∈(0,1),

u(0) =u(1) =u0(0) =u0(1) = 0, (1) where α∈(3,4] is a real number, Dα0+ is the standard Riemann-Liouville derivative, f ∈ C([0,1]×[0,∞),[0,∞)), and h ∈ C(0,1)∩L(0,1) is nonnegative and may be singular at t= 0 and/ort= 1. We use fixed point index theory to establish our main results based on a priori estimates achieved by developing some spectral properties of associated linear integral operators.

Fractional differential equations can describe many phenomena in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, electromagnetics, etc. This explains why many authors have studied existence and mul- tiplicity questions of solutions (or positive solutions) of nonlinear fractional differential

Mathematics Subject Classifications: 34B16, 34B18, 47H07, 47H11, 45M20.

Department of Mathematics, Qingdao Technological University, Qingdao, Shandong Province, PR China.

259

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equation, see, for example, [1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 13] and references therein. It is of interest to note that the Riemann-Liouville fractional derivative is not suitable for nonzero boundary value conditions, see [11, 13].

By means of the Schauder fixed point theorem and fixed point index theory, Bai [1]

discussed the existence of positive solutions for the fractional boundary value problem (Dα0+u(t) +f(t, u(t)) = 0, t∈(0,1),

u(0) = 0, βu(η) =u(1), (2)

whereα∈(1,2],Dα0+is the Riemann-Liouville fractional derivative,η∈(0,1),βηα−1∈ (0,1), andf ∈C([0,1]×[0,∞),[0,∞)) is sublinear. It should be remarked that our nonlinearity f here, unlike the f in [1], may be both sublinear and superlinear. Our first theorem involves the existence of at least one positive solution for (1) with f growing superlinearly, thereby complementing the results in [1]. We then establish two existence results of twin positive solutions for (1), two results that essentially improve and extend the corresponding ones in [12] (see REMARK 1).

2 Preliminaries

The Riemann-Liouville fractional derivativeDα0+ is defined by Dα0+y(t) =Γ(n−α)1 dtdnRt

0

y(s)ds (t−s)αn+1,

where Γ is the gamma function andn= [α] + 1. For more details of fractional calculus, we refer the reader to the recent literature such as [1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 13].

Let

E=C[0,1], kuk= max

t∈[0,1]|u(t)|, P ={u∈E:u(t)≥0,∀t∈[0,1]}.

Then (E,k · k) is a real Banach space and P a cone onE. We denoteBρ ={u∈E : kuk< ρ}forρ >0 in the sequel.

LEMMA 1 ([12, Lemma 2.3]). Given φ ∈ C(0,1)∩L(0,1) and 3 < α ≤ 4, the unique solution of the fractional boundary value problem

(Dα0+u(t) =φ(t), t∈(0,1), u(0) =u(1) =u0(0) =u0(1) = 0, is represented by u(t) =R1

0 G(t, s)φ(s)ds, where G(t, s) =

((t−s)α1

+(1−s)α2tα2[(s−t)+(α−2)(1−t)s]

Γ(α) , 0≤s≤t≤1,

tα2(1−s)α2[(s−t)+(α−2)(1−t)s]

Γ(α) , 0≤t≤s≤1. (3)

It is easy to verify that the Green’s functionG∈ C([0,1]×[0,1],[0,∞)) satisfies the following relations(see [12, Lemma 2.4])

(α−2)tα−2(1−t)2s2(1−s)α−2≤Γ(α)G(t, s)≤m0s2(1−s)α−2 (4)

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and

(α−2)q(t)k(s)≤Γ(α)G(t, s)≤m0k(s) (5) for all 0≤t, s≤1, whereq(t) =tα−2(1−t)2,k(s) =s2(1−s)α−2 and

m0= max{α−1,(α−2)2}. (6)

LEMMA 2 ([5, p.314]). Suppose A: P → P is a completely continuous operator and has no fixed points on ∂Bρ∩P. Then the following are true:

1. IfkAuk ≤ kukfor allu∈∂Bρ∩P, then i(A, Bρ∩P, P) = 1, where iis the fixed point index onP .

2. IfkAuk ≥ kukfor allu∈∂Bρ∩P, theni(A, Bρ∩P, P) = 0.

LEMMA 3 ([5, p.144]). Suppose Ω⊂E is a bounded open set andA: Ω∩P →P is a completely continuous operator. If there existsu0∈P\ {0}such that

u−Au6=λu0,∀λ≥0, u∈∂Ω∩P, then i(A,Ω∩P, P) = 0 .

LEMMA 4. ([5, p.164]) Let Ω⊂E be a bounded open set with 0∈ Ω. Suppose A: Ω∩P →P is a completely continuous operator and has no fixed points on∂Ω∩P. Ifu6=λAu,∀u∈∂Ω∩P,0≤λ≤1, then i(A,Ω∩P, P) = 1.

We assume the following conditions throughout this paper.

(H1)f ∈C([0,1]×[0,∞),[0,∞)).

(H2) h ∈ L(0,1)∩C(0,1) is nonnegative and does not vanish identically on any subinterval of (0,1).

Define two operatorsA andT by (Au)(t) =

Z 1 0

G(t, s)h(s)f(s, u(s))ds, u∈P and

(T u)(t) = Z 1

0

G(t, s)h(s)u(s)ds, u∈P.

Note that (H1) and (H2) implyA :P → P is a completely continuous operator and T : P → P is a completely continuous, linear, positive operator. A consequence of Lemma 1 is that u ∈ P is a positive solution of (1) if and only if u ∈ P \ {0} is a fixed point of A. Moreover, it is easy to prove that the spectral radius of T, denoted byr(T), is positive. Now the well-known Krein-Rutman theorem [9] asserts that there exist two functions ϕ∈P\ {0} andψ∈L(0,1)\{0} withψ(x)≥0 for which

Z 1 0

G(t, s)h(s)ϕ(s)ds=r(T)ϕ(t), Z 1

0

G(t, s)h(s)ψ(t)dt=r(T)ψ(s), Z 1

0

ψ(t)dt= 1.

(7) Put

P0={u∈P : Z 1

0

ψ(t)u(t)dt≥ωkuk},

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where ψ(t) is determined by (7) and ω = α−2m

0

R1

0 q(t)ψ(t)dt >0. Clearly,P0 is also a cone onE. The following is a result obtained by observing relations (5).

LEMMA 5. A(P)⊂P0.

In addition to (H1) and (H2), we need the following hypotheses onf. (H3) lim infu→0+f(t,u)

u > λ1 uniformly with respect to t ∈ [0,1], where λ1 = 1/r(T)>0.

(H4) lim supu→+∞f(t,u)u < λ1 uniformly with respect tot∈[0,1].

(H5) lim infu→+∞ f(t,u)

u > λ1 uniformly with respect tot∈[0,1].

(H6) lim supu→0+f(t,u)u < λ1 uniformly with respect tot∈[0,1].

(H7) There is ρ > 0 such that the inequality f(t, u) < ηΓ(α)ρm

0 holds whenever u∈[0, ρ] andt∈[0,1] ,m0>0 being defined in (6).

(H8) There are ρ > 0 and σ ∈ (0,α−2α ) such that f(t, u) > η(α−2)q(Γ(α)ρα2 α ) holds wheneveru∈[θρ, ρ] andt∈[σ,1−σ],θandηbeing defined byθ= α−2m

0 min{q(σ), q(1−

σ)}>0 andη=R1

0 k(s)h(s)ds−1

>0.

REMARK 1. Some simple computations show the following estimates:

Γ(α) m0

Z 1 0

k(s)ds −1

≤λ1 = (r(T))−1 ≤ m0

α−2

t∈[0,1]max Z 1

0

G(t, s)q(s)ds −1

, from which we obtainM ≤λ1≤N ≤N ,e where M, N ,Ne are defined in [12, Section 3]. Notice that (H3) and (H5) considerably weaken (A1), and that (H4) and (H6) considerably (A2), where (A1) and (A2) are formulated in [12]. This means our main results, even in the case of h being nonsingular, essentially extend and improve the corresponding ones in the literature.

3 Main results

First we have the following.

THEOREM 1. Suppose that (H1), (H2), (H5) and (H6) are satisfied, then (1) has at least one positive solution.

PROOF. By (H5), there existε >0 andb >0 such thatf(t, u)≥(λ1+ε)u−bfor allu≥0 andt∈[0,1]. This implies

(Au)(t)≥(λ1+ε) Z 1

0

G(t, s)h(s)u(s)ds−b Z 1

0

G(t, s)h(s)ds (8) for allu∈P. LetM1={u∈P :u=Au+λϕ, λ≥0}, whereϕ∈P is determined by (7). We shall prove that M1 is bounded. Indeed, ifu∈M1, then we haveu≥Auby definition. This together with (8) leads to

u(t)≥(λ1+ε) Z 1

0

G(t, s)h(s)u(s)ds−b Z 1

0

G(t, s)h(s)ds.

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Multiply by ψ(t) on both sides of the above and integrate over [0,1] and use (7) to obtain

R1

0 ψ(t)u(t)dt≥(λ1+ε)λ−11 R1

0 ψ(t)u(t)dt−bλ−11 , so that R1

0 ψ(t)u(t)dt≤ bε for allu∈M1. Note we have M1 ⊂P0 by Lemma 5. This together with the preceding inequality implies kuk ≤ (εω)−1b for allu ∈M1, which establishes the boundedness ofM1, as required. TakingR >(εω)−1b, we obtain

u6=Au+λϕ, ∀u∈∂BR∩P, λ≥0.

Now Lemma 3 yields

i(A, BR∩P, P) = 0. (9)

By (H6), there exist r ∈ (0, R) andε ∈ (0, λ1) such that f(t, u) ≤ (λ1−ε)u for all u∈[0, r] andt∈[0,1]. This implies

(Au)(t)≤(λ1−ε) Z 1

0

G(t, s)h(s)u(s)ds (10)

for allu∈Br∩P. Now we claim

u6=µAu, ∀u∈∂Br∩P, 0≤µ≤1. (11) Indeed, if there exist u0 ∈∂Br∩P and µ0 ∈[0,1] for which u00Au0, then this together with (10) leads tou0(t)≤(λ1−ε)R1

0 G(t, s)h(s)u0(s)ds. Multiply byψ(t) on both sides of the preceding inequality and integrate over [0,1] and use (7) to obtain

Z 1 0

ψ(t)u0(t)dt≤λ1−ε λ1

Z 1 0

ψ(t)u0(t)dt, so thatR1

0ψ(t)u0(t)dt= 0, whenceu0(t)≡0, contradictingu0∈∂Br∩P. As a result, (11) is true and we have by Lemma 4

i(A, Br∩P, P) = 1. (12)

Now (9) and (12) combined imply

i(A,(BR\Br)∩P, P) =i(A, BR∩P, P)−i(A, Br∩P, P) =−1.

Hence the operatorAhas at least one fixed point on (BR\Br)∩P. Therefore (1) has at least one positive solution, which completes the proof.

THEOREM 2. Suppose that (H1)-(H3), (H5) and (H7) are satisfied, then (1) has at least two positive solutions.

PROOF. By (H7), we have kAuk= max

t∈[0,1]

Z 1 0

G(t, s)h(s)f(s, u(s))ds <

Z 1 0

m0

Γ(α)k(s)h(s)ηΓ(α)ρ M0

ds=kuk

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for allu∈∂Bρ∩P. Now Lemma 2 yields

i(A, Bρ∩P, P) = 1. (13)

On the other hand, in view of (H5), we may takeR > ρso that (9) holds (see the proof of Theorem 1). By (H3), there exist r∈(0, ρ) andε >0 such thatf(t, u)≥(λ1+ε)u, for allu∈[0, r] andt∈[0,1]. This implies

(Au)(t)≥(λ1+ε) Z 1

0

G(t, s)h(s)u(s)ds (14)

for allu∈Br∩P. Now we claim

u−Au6=µϕ,∀u∈∂Br∩P, µ≥0, (15) whereϕis determined by (7). Indeed, if the claim is false, then there existu1∈∂Br∩P andµ1≥0 such thatu1−Au11ϕand thusu1≥Au1. Combining the last inequality with (14) (with ureplaced byu1), we obtain

u1(t)≥(λ1+ε) Z 1

0

G(t, s)h(s)u1(s)ds.

Multiply by ψ(t) on both sides of the above and integrate over [0,1] and use (7) to

obtain Z 1

0

u1(t)ψ(t)dt≥(λ1+ε)λ−11 Z 1

0

u1(t)ψ(t)dt, so thatR1

0 u1(t)ψ(t)dt= 0, whenceu1(t)≡0, contradictingu1∈∂Br∩P. As a result, (15) is true, as claimed. Now Lemma 3 yields

i(A, Br∩P, P) = 0. (16)

Combining (9), (13) and (16), we arrive at

i(A,(BR\Bρ)∩P, P) = 0−1 =−1 and

i(A,(Bρ\Br)∩P, P) = 1−0 = 1.

Consequently,Ahas at least two fixed points, with one on (BR\Bρ)∩P and the other on (Bρ\Br)∩P. Therefore (1) has at least two positive solutions, which completes the proof.

To prove Theorem 3 below, we need an extra coneP1, which is defined by P1={u∈P :u(t)≥θkuk,∀t∈[σ,1−σ]},

where θ= α−2m

0 min{q(σ), q(1−σ)}.We have the following Claim 1. A(P)⊂P1.

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PROOF. On the one hand,u∈P implieskAuk ≤ Γ(α)1 R1

0 m0k(s)h(s)f(s, u(s))ds.

On the other hand,u∈P andt∈[σ,1−σ] imply (Au)(t)≥ (α−2)q(t)

m0Γ(α) Z 1

0

m0k(s)h(s)f(s, u(s))ds≥ θ Γ(α)

Z 1 0

m0k(s)h(s)f(s, u(s))ds and thus (Au)(t)≥θkAukfor allu∈P andt∈[σ,1−σ]. This impliesA(P)⊂P1, as claimed.

THEOREM 3. Suppose that (H1), (H2), (H4), (H6) and (H8) are satisfied, then (1) has at least two positive solutions.

PROOF. RecallA(P)⊂P1. By (H8), we have kAuk = max

0≤t≤1(Au)(t)≥ max

t∈[σ,1−σ](Au)(t)

= max

t∈[σ,1−σ]

Z 1 0

G(t, s)h(s)f(s, u(s))ds

> max

t∈[σ,1−σ]

Z 1 0

(α−2)q(t)k(s)

Γ(α) h(s)f(s, u(s))ds

= (α−2)q(α−2α ) Γ(α)

Z 1 0

k(s)h(s) Γ(α)ρ η(α−2)q(α−2α )ds

= kuk, for allu∈∂Bρ∩P, and by Lemma 2

i(A, Bρ∩P, P) = 0. (17)

On the other hand, in view of (H6) , we may taker∈(0, ρ) so that (12) holds (see the proof of Theorem 1). In addition, by (H4), there existε∈(0, λ1) andm >0 such that f(t, u)≤(λ1−ε)u+m for allu≥0 andt∈[0,1]. Let

M2={u∈P : u=µAu, 0≤µ≤1}. (18) We shall prove thatM2is bounded. Indeed, ifu∈M2, then, by definition, we have for someµ∈[0,1]

u(t) =µ(Au)(t)≤R1

0 G(t, s)h(s)f(s, u(s))ds

≤R1

0 G(t, s)h(s)((λ1−ε)u(s) +m)ds

= (λ1−ε)(T u)(t) +u0(t),

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u0 ∈ P \ {0} being defined by u0(t) = mR1

0 G(t, s)h(s)ds. Notice r((λ1−ε)T) <1.

This implies the inverse operator of I−(λ1−ε)T exists and equals

(I−(λ1−ε)T)−1=I+ (λ1−ε)T+ (λ1−ε)2T2+. . .+ (λ1−ε)nTn+. . . , from which we obtain (I−(λ1−ε)T)−1(P) ⊂ P. Applying this to (19) gives u ≤ (I−(λ1−ε)T)−1u0 for all u∈M2. This proves the boundedness ofM2, as required.

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Choosing R >sup{kuk:u∈M2} and R > ρ, we haveu6=λAu for all u∈∂BR∩P and λ∈[0,1]. Now Lemma 4 yields

i(A, BR∩P, P) = 1. (20)

Combining (12), (17) and (20), we arrive at

i(A,(BR\Bρ)∩P, P) = 1−0 = 1, i(A,(Bρ\Br)∩P, P) = 0−1 =−1.

Hence A has at least two fixed points, with one on (BR\Bρ)∩P and the other on (Bρ\Br)∩P. Therefore (1) has at least two positive solutions, which completes the proof.

References

[1] Z. Bai, On positive solutions of a nonlocal fractional boundary value problem, Nonlinear Analysis, 72(2010), 916–924.

[2] Z. Bai and H. L¨u, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl., 311(2005), 495–505.

[3] A. Babakhani and V. D. Gejji, Existence of positive solutions of nonlinear frac- tional differential equations, J. Math. Anal. Appl., 278(2003), 434–442.

[4] D. Delbosco and L. Rodino, Existence and uniqueness for a fractional differential equation, J. Math. Anal. Appl., 204(1996), 609–625.

[5] D. Guo, Nonlinear Functional Analysis, Jinan: Science and Technology Press of Shandong, 1985 (in Chinese).

[6] D. Jiang and C. Yuan, The positive properties of the Green function for Dirichlet- type boundary value problems of nonlinear fractional differential equations and its application, Nonlinear Analysis, 72(2010), 710–719.

[7] A. A. Kilbas and J. J. Trujillo, Differential equations of fractional order: methods, results and problems-I, Appl. Anal., 78(2001), 153–192.

[8] A. A. Kilbas and J. J. Trujillo, Differential equations of fractional order: methods, results and problems-II, Appl. Anal. 81 (2002), 435–493.

[9] M. G. Krein and M. A. Rutman, Linear operators leaving invariant a cone in a Banach space, Transl. Amer. Math. Soc., 10(1962), 199–325.

[10] I. Podlubny, Fractional Differential Equations, Mathematics in Science and Engi- neering, Academic Press, New York, London, Toronto, 1999.

[11] X. Su and S. Zhang, Solutions to boundary-value problems for nonlinear differ- ential equations of fractional order, Electronic Journal of Differential Equations, 2009(2009), No. 26, pp. 1-15.

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[12] X. Xu, D. Jiang and C. Yuan, Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equation, Nonlinear Analysis, 71(2009), 4676–4688.

[13] S. Zhang, Positive solutions for boundary-value problems of nonlinear frac- tional differential equations, Electronic Journal of Differential Equations, Vol.

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